Lesson type: learning new material.
Lesson objectives:
- expansion and deepening of students’ understanding of problems solved using arithmetic progression; organization search activity students when deriving the formula for the sum of the first n terms of an arithmetic progression;
- developing the ability to independently acquire new knowledge and use already acquired knowledge to achieve a given task;
- developing the desire and need to generalize the facts obtained, developing independence.
Tasks:
- summarize and systematize existing knowledge on the topic “Arithmetic progression”;
- derive formulas for calculating the sum of the first n terms of an arithmetic progression;
- teach how to apply the obtained formulas when solving various problems;
- draw students' attention to the procedure for finding the value of a numerical expression.
Equipment:
- cards with tasks for working in groups and pairs;
- evaluation paper;
- presentation“Arithmetic progression.”
I. Updating of basic knowledge.
1. Independent work in pairs.
1st option:
Define arithmetic progression. Write down a recurrence formula that defines an arithmetic progression. Please provide an example of an arithmetic progression and indicate its difference.
2nd option:
Write down the formula for the nth term of an arithmetic progression. Find the 100th term of the arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students on the back of the board are preparing answers to the same questions.
Students evaluate their partner's work by checking them on the board. (Sheets with answers are handed in.)
2. Game moment.
Exercise 1.
Teacher. I thought of some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th term of this progression. (1, 3, 5, 7, 9, 11, 13, 15…)
Questions from students.
- What is the sixth term of the progression and what is the difference?
- What is the eighth term of the progression and what is the difference?
If there are no more questions, then the teacher can stimulate them - a “ban” on d (difference), that is, it is not allowed to ask what the difference is equal to. You can ask questions: what is the 6th term of the progression equal to and what is the 8th term of the progression equal to?
Task 2.
There are 20 numbers written on the board: 1, 4, 7 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58.
The teacher stands with his back to the board. Students call out the number, and the teacher instantly calls out the number itself. Explain how I can do this?
The teacher remembers the formula for the nth term a n = 3n – 2 and, substituting the specified values n, finds the corresponding values a n.
II. Setting a learning task.
I propose to solve an ancient problem dating back to the 2nd millennium BC, found in Egyptian papyri.
Task:“Let it be said to you: divide 10 measures of barley among 10 people, the difference between each person and his neighbor is 1/8 of the measure.”
- How is this problem related to the topic arithmetic progression? (Each next person receives 1/8 of the measure more, which means the difference is d=1/8, 10 people, which means n=10.)
- What do you think the number 10 measures means? (Sum of all terms of the progression.)
- What else do you need to know to make it easy and simple to divide the barley according to the conditions of the problem? (First term of progression.)
Lesson Objective– obtaining the dependence of the sum of the terms of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.
Before we deduce the formula, let's look at how the ancient Egyptians solved the problem.
And they solved it as follows:
1) 10 measures: 10 = 1 measure – average share;
2) 1 measure ∙ = 2 measures – doubled average share.
Doubled average share is the sum of the shares of the 5th and 6th person.
3) 2 measures – 1/8 measures = 1 7/8 measures – double the share of the fifth person.
4) 1 7/8: 2 = 5/16 – fraction of a fifth; and so on, you can find the share of each previous and subsequent person.
We get the sequence:
III. Solving the problem.
1. Work in groups
Group I: Find the sum of 20 consecutive natural numbers: S 20 =(20+1)∙10 =210.
IN general view
II group: Find the sum of natural numbers from 1 to 100 (The Legend of Little Gauss).
S 100 = (1+100)∙50 = 5050
Conclusion:
III group: Find the sum of natural numbers from 1 to 21.
Solution: 1+21=2+20=3+19=4+18…
Conclusion:
IV group: Find the sum of natural numbers from 1 to 101.
Conclusion:
This method of solving the problems considered is called the “Gauss Method”.
2. Each group presents the solution to the problem on the board.
3. Generalization of the proposed solutions for an arbitrary arithmetic progression:
a 1, a 2, a 3,…, a n-2, a n-1, a n.
S n =a 1 + a 2 + a 3 + a 4 +…+ a n-3 + a n-2 + a n-1 + a n.
Let's find this sum using similar reasoning:
4. Have we solved the problem?(Yes.)
IV. Primary understanding and application of the obtained formulas when solving problems.
1. Checking the solution to an ancient problem using the formula.
2. Application of the formula in solving various problems.
3. Exercises to develop the ability to apply formulas when solving problems.
A) No. 613
Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …, 1500
Find: S 1500
Solution: , a 1 = 1, and 1500 = 1500,
B) Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …
S n = 210
Find: n
Solution:
V. Independent work with mutual verification.
Denis started working as a courier. In the first month his salary was 200 rubles, in each subsequent month it increased by 30 rubles. How much did he earn in total in a year?
Given: ( a n) – arithmetic progression;
a 1 = 200, d=30, n=12
Find: S 12
Solution:
Answer: Denis received 4380 rubles for the year.
VI. Homework instruction.
- Section 4.3 – learn the derivation of the formula.
- №№ 585, 623 .
- Create a problem that can be solved using the formula for the sum of the first n terms of an arithmetic progression.
VII. Summing up the lesson.
1. Score sheet
2. Continue the sentences
- Today in class I learned...
- Formulas learned...
- I believe that …
3. Can you find the sum of numbers from 1 to 500? What method will you use to solve this problem?
Bibliography.
1. Algebra, 9th grade. Tutorial for educational institutions. Ed. G.V. Dorofeeva. M.: “Enlightenment”, 2009.
When studying algebra in secondary school(9th grade) one of important topics is the study of number sequences, which include progressions - geometric and arithmetic. In this article we will look at an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to define the progression in question, as well as provide the basic formulas that will be used later in solving problems.
An arithmetic or algebraic progression is a set of ordered rational numbers, each term of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.
Let's give an example. The following sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the type of progression under consideration, since the difference for it is not constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important Formulas
Let us now present the basic formulas that will be needed to solve problems using arithmetic progression. Let us denote by the symbol a n nth term sequences where n is an integer. We denote the difference by the Latin letter d. Then the following expressions are valid:
- To determine the value of the nth term, the following formula is suitable: a n = (n-1)*d+a 1 .
- To determine the sum of the first n terms: S n = (a n +a 1)*n/2.
To understand any examples of arithmetic progression with solutions in 9th grade, it is enough to remember these two formulas, since any problems of the type under consideration are based on their use. You should also remember that the progression difference is determined by the formula: d = a n - a n-1.
Example #1: finding an unknown member
Let's give a simple example of an arithmetic progression and the formulas that need to be used to solve it.
Let the sequence 10, 8, 6, 4, ... be given, you need to find five terms in it.
From the conditions of the problem it already follows that the first 4 terms are known. The fifth can be defined in two ways:
- Let's first calculate the difference. We have: d = 8 - 10 = -2. Similarly, you could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d = a n - a n-1, then d = a 5 - a 4, from which we get: a 5 = a 4 + d. We substitute the known values: a 5 = 4 + (-2) = 2.
- The second method also requires knowledge of the difference of the progression in question, so you first need to determine it as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n = (n - 1) * d + a 1 = (n - 1) * (-2) + 10 = 12 - 2*n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.
As you can see, both solutions led to the same result. Note that in this example the progression difference d is a negative value. Such sequences are called decreasing, since each next term is less than the previous one.
Example #2: progression difference
Now let’s complicate the task a little, let’s give an example of how
It is known that in some the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.
Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . Let's substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression you can easily calculate the difference: d = (18 - 6) /6 = 2. Thus, we have answered the first part of the problem.
To restore the sequence to the 7th term, you should use the definition of an algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d, and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16, a 7 = 18.
Example No. 3: drawing up a progression
Let's complicate the problem even more. Now we need to answer the question of how to find an arithmetic progression. The following example can be given: two numbers are given, for example - 4 and 5. It is necessary to create an algebraic progression so that three more terms are placed between these.
Before you start solving this problem, you need to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we move on to the problem, which is similar to the previous one. Again, for the nth term we use the formula, we get: a 5 = a 1 + 4 * d. From: d = (a 5 - a 1)/4 = (5 - (-4)) / 4 = 2.25. What we got here is not an integer value of the difference, but it is a rational number, so the formulas for the algebraic progression remain the same.
Now let's add the found difference to a 1 and restore the missing terms of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 = 2.75 + 2.25 = 5, which coincided with the conditions of the problem.
Example No. 4: first term of progression
Let's continue to give examples of arithmetic progression with solutions. In all previous problems, the first number of the algebraic progression was known. Now let's consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find which number this sequence begins with.
The formulas used so far assume knowledge of a 1 and d. In the problem statement, nothing is known about these numbers. Nevertheless, we will write down expressions for each term about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We received two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The easiest way to solve this system is to express a 1 in each equation and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 0.464 (only 3 decimal places are given).
Knowing d, you can use any of the 2 expressions above for a 1. For example, first: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.
If you have doubts about the result obtained, you can check it, for example, determine the 43rd term of the progression, which is specified in the condition. We get: a 43 = a 1 + 42 * d = 56.496 + 42 * (- 0.464) = 37.008. The small error is due to the fact that rounding to thousandths was used in the calculations.
Example No. 5: amount
Now let's look at several examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to the development of computer technology, it is possible to solve this problem, that is, add all the numbers sequentially, which Calculating machine will do as soon as the person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is equal to 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is interesting to note that this problem is called “Gaussian” because at the beginning of the 18th century the famous German, still only 10 years old, was able to solve it in his head in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add the numbers at the ends of the sequence in pairs, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer it is enough to multiply 50 by 101.
Example No. 6: sum of terms from n to m
Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its terms from 8 to 14 will be equal to.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them sequentially. Since there are few terms, this method is not quite labor-intensive. Nevertheless, it is proposed to solve this problem using a second method, which is more universal.
The idea is to obtain a formula for the sum of the algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:
- S m = m * (a m + a 1) / 2.
- S n = n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2nd sum includes the first. The last conclusion means that if we take the difference between these sums and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), we will obtain the necessary answer to the problem. We have: S mn = S n - S m + a m =n * (a 1 + a n) / 2 - m *(a 1 + a m)/2 + a m = a 1 * (n - m) / 2 + a n * n/2 + a m * (1- m/2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d *(3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.
As can be seen from the above solutions, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before starting to solve any of these problems, it is recommended that you carefully read the condition, clearly understand what you need to find, and only then proceed with the solution.
Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the likelihood of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and divide the overall problem into separate subtasks (in this case, first find the terms a n and a m).
If you have doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. We found out how to find an arithmetic progression. If you figure it out, it's not that difficult.
What the main point formulas?
This formula allows you to find any BY HIS NUMBER " n" .
Of course, you also need to know the first term a 1 and progression difference d, well, without these parameters you can’t write down a specific progression.
Memorizing (or cribing) this formula is not enough. You need to understand its essence and apply the formula in various problems. And also not to forget at the right moment, yes...) How not forget- I don't know. And here how to remember If necessary, I will definitely advise you. For those who complete the lesson to the end.)
So, let's look at the formula for the nth term of an arithmetic progression.
What is a formula in general? By the way, take a look if you haven’t read it. Everything is simple there. It remains to figure out what it is nth term.
Progression in general can be written as a series of numbers:
a 1, a 2, a 3, a 4, a 5, .....
a 1- denotes the first term of an arithmetic progression, a 3- third member, a 4- the fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - s a 120.
How can we define it in general terms? any term of an arithmetic progression, with any number? Very simple! Like this:
a n
That's what it is nth term of an arithmetic progression. The letter n hides all the member numbers at once: 1, 2, 3, 4, and so on.
And what does such a record give us? Just think, instead of a number they wrote down a letter...
This notation gives us a powerful tool for working with arithmetic progression. Using the notation a n, we can quickly find any member any arithmetic progression. And solve a bunch of other progression problems. You'll see for yourself further.
In the formula for the nth term of an arithmetic progression:
| a n = a 1 + (n-1)d |
a 1- the first term of an arithmetic progression;
n- member number.
The formula connects the key parameters of any progression: a n ; a 1 ; d And n. All progression problems revolve around these parameters.
The nth term formula can also be used to write a specific progression. For example, the problem may say that the progression is specified by the condition:
a n = 5 + (n-1) 2.
Such a problem can be a dead end... There is neither a series nor a difference... But, comparing the condition with the formula, it is easy to understand that in this progression a 1 =5, and d=2.
And it can be even worse!) If we take the same condition: a n = 5 + (n-1) 2, Yes, open the parentheses and bring similar ones? We get a new formula:
a n = 3 + 2n.
This Just not general, but for a specific progression. This is where the pitfall lurks. Some people think that the first term is a three. Although in reality the first term is five... A little lower we will work with such a modified formula.
In progression problems there is another notation - a n+1. This is, as you guessed, the “n plus first” term of the progression. Its meaning is simple and harmless.) This is a member of the progression whose number is greater than number n by one. For example, if in some problem we take a n fifth term then a n+1 will be the sixth member. Etc.
Most often the designation a n+1 found in recurrence formulas. Don't be afraid of this scary word!) This is just a way of expressing a member of an arithmetic progression through the previous one. Let's say we are given an arithmetic progression in this form, using a recurrent formula:
a n+1 = a n +3
a 2 = a 1 + 3 = 5+3 = 8
a 3 = a 2 + 3 = 8+3 = 11
The fourth - through the third, the fifth - through the fourth, and so on. How can we immediately count, say, the twentieth term? a 20? But there’s no way!) Until we find out the 19th term, we can’t count the 20th. This is the fundamental difference between the recurrent formula and the formula of the nth term. Recurrent works only through previous term, and the formula of the nth term is through first and allows straightaway find any member by its number. Without calculating the entire series of numbers in order.
In an arithmetic progression, it is easy to turn a recurrent formula into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in its usual form, and work with it. Such tasks are often encountered in the State Academy of Sciences.
Application of the formula for the nth term of an arithmetic progression.
First, let's look at the direct application of the formula. At the end of the previous lesson there was a problem:
An arithmetic progression (a n) is given. Find a 121 if a 1 =3 and d=1/6.
This problem can be solved without any formulas, simply based on the meaning of an arithmetic progression. Add and add... An hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) Let's decide.
The conditions provide all the data for using the formula: a 1 =3, d=1/6. It remains to figure out what is equal n. No problem! We need to find a 121. So we write:
Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be ours n. This is the meaning n= 121 we will substitute further into the formula, in brackets. We substitute all the numbers into the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3+20 = 23
That's it. Just as quickly one could find the five hundred and tenth term, and the thousand and third, any one. We put instead n the desired number in the index of the letter " a" and in brackets, and we count.
Let me remind you the point: this formula allows you to find any arithmetic progression term BY HIS NUMBER " n" .
Let's solve the problem in a more cunning way. Let us come across the following problem:
Find the first term of the arithmetic progression (a n), if a 17 =-2; d=-0.5.
If you have any difficulties, I will tell you the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Write down with your hands, right in your notebook:
| a n = a 1 + (n-1)d |
And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member... Is that it? If you think that’s it, then you won’t solve the problem, yes...
We still have a number n! In condition a 17 =-2 hidden two parameters. This is both the value of the seventeenth term (-2) and its number (17). Those. n=17. This “trifle” often slips past the head, and without it, (without the “trifle”, not the head!) the problem cannot be solved. Although... and without a head too.)
Now we can simply stupidly substitute our data into the formula:
a 17 = a 1 + (17-1)·(-0.5)
Oh yes, a 17 we know it's -2. Okay, let's substitute:
-2 = a 1 + (17-1)·(-0.5)
That's basically all. It remains to express the first term of the arithmetic progression from the formula and calculate it. The answer will be: a 1 = 6.
This technique - writing down a formula and simply substituting known data - is a great help in simple tasks. Well, of course, you must be able to express a variable from a formula, but what to do!? Without this skill, mathematics may not be studied at all...
Another popular puzzle:
Find the difference of the arithmetic progression (a n), if a 1 =2; a 15 =12.
What are we doing? You will be surprised, we are writing the formula!)
| a n = a 1 + (n-1)d |
Let's consider what we know: a 1 =2; a 15 =12; and (I’ll especially highlight!) n=15. Feel free to substitute this into the formula:
12=2 + (15-1)d
We do the arithmetic.)
12=2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, the tasks for a n, a 1 And d decided. All that remains is to learn how to find the number:
The number 99 is a member of the arithmetic progression (a n), where a 1 =12; d=3. Find this member's number.
We substitute the quantities known to us into the formula of the nth term:
a n = 12 + (n-1) 3
At first glance, there are two unknown quantities here: a n and n. But a n- this is some member of the progression with a number n...And we know this member of the progression! It's 99. We don't know its number. n, So this number is what you need to find. We substitute the term of the progression 99 into the formula:
99 = 12 + (n-1) 3
We express from the formula n, we think. We get the answer: n=30.
And now a problem on the same topic, but more creative):
Determine whether the number 117 is a member of the arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Let's write the formula again. What, there are no parameters? Hm... Why are we given eyes?) Do we see the first term of the progression? We see. This is -3.6. You can safely write: a 1 = -3.6. Difference d Can you tell from the series? It’s easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
So, we did the simplest thing. It remains to deal with the unknown number n and the incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know... What to do!? Well, what to do, what to do... Turn on Creative skills!)
We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes, yes!)) and substitute our numbers:
117 = -3.6 + (n-1) 1.2
Again we express from the formulan, we count and get:
Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion can we draw? Yes! Number 117 is not member of our progression. It is somewhere between the one hundred and first and one hundred and second terms. If the number turned out natural, i.e. is a positive integer, then the number would be a member of the progression with the number found. And in our case, the answer to the problem will be: No.
A task based on a real version of the GIA:
An arithmetic progression is given by the condition:
a n = -4 + 6.8n
Find the first and tenth terms of the progression.
Here the progression is set in an unusual way. Some kind of formula... It happens.) However, this formula (as I wrote above) - also the formula for the nth term of an arithmetic progression! She also allows find any member of the progression by its number.
We are looking for the first member. The one who thinks. that the first term is minus four is fatally mistaken!) Because the formula in the problem is modified. The first term of the arithmetic progression in it hidden. It’s okay, we’ll find it now.)
Just as in previous problems, we substitute n=1 into this formula:
a 1 = -4 + 6.8 1 = 2.8
Here! The first term is 2.8, not -4!
We look for the tenth term in the same way:
a 10 = -4 + 6.8 10 = 64
That's it.
And now, for those who have read to these lines, the promised bonus.)
Suppose, in a difficult combat situation, State Examination or Unified State Examination, you forgot useful formula nth term of an arithmetic progression. I remember something, but somehow uncertainly... Or n there, or n+1, or n-1... How to be!?
Calm! This formula is easy to derive. It’s not very strict, but it’s definitely enough for confidence and the right decision!) To make a conclusion, it’s enough to remember the elementary meaning of an arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
Draw a number line and mark the first one on it. second, third, etc. members. And we note the difference d between members. Like this:
We look at the picture and think: what does the second term equal? Second one d:
a 2 =a 1 + 1 d
What is the third term? Third term equals first term plus two d.
a 3 =a 1 + 2 d
Do you get it? It’s not for nothing that I highlight some words in bold. Okay, one more step).
What is the fourth term? Fourth term equals first term plus three d.
a 4 =a 1 + 3 d
It's time to realize that the number of gaps, i.e. d, Always one less than the number of the member you are looking for n. That is, to the number n, number of spaces will n-1. Therefore, the formula will be (without variations!):
| a n = a 1 + (n-1)d |
In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it’s difficult to draw a picture, then... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't insert a picture into the equation...
Tasks for independent solution.
To warm up:
1. In arithmetic progression (a n) a 2 =3; a 5 =5.1. Find a 3 .
Hint: according to the picture, the problem can be solved in 20 seconds... According to the formula, it turns out more difficult. But for mastering the formula, it’s more useful.) In Section 555, this problem is solved using both the picture and the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 =19.1; a 236 =49, 3. Find a 3 .
What, you don’t want to draw a picture?) Of course! Better according to the formula, yes...
3. The arithmetic progression is given by the condition:a 1 = -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.
In this task, the progression is specified in a recurrent manner. But counting to the one hundred and twenty-fifth term... Not everyone is capable of such a feat.) But the formula of the nth term is within the power of everyone!
4. Given an arithmetic progression (a n):
-148; -143,8; -139,6; -135,4, .....
Find the number of the smallest positive term of the progression.
5. According to the conditions of task 4, find the sum of the smallest positive and largest negative terms of the progression.
6. The product of the fifth and twelfth terms of an increasing arithmetic progression is equal to -2.5, and the sum of the third and eleventh terms is equal to zero. Find a 14 .
Not the easiest task, yes...) The “fingertip” method won’t work here. You will have to write formulas and solve equations.
Answers (in disarray):
3,7; 3,5; 2,2; 37; 2,7; 56,5
Happened? It's nice!)
Not everything works out? Happens. By the way, there is one subtle point in the last task. Care will be required when reading the problem. And logic.
The solution to all these problems is discussed in detail in Section 555. And the element of fantasy for the fourth, and the subtle point for the sixth, and general approaches for solving any problems involving the formula of the nth term - everything is described. I recommend.
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By the way, I have a couple more interesting sites for you.)
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You can get acquainted with functions and derivatives.
Mathematics has its own beauty, just like painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common tasks in entrance examinations in mathematics are problems related to the concept of arithmetic progression. To successfully solve such problems, you must have a good knowledge of the properties of arithmetic progression and have certain skills in their application.
Let us first recall the basic properties of an arithmetic progression and present the most important formulas, related to this concept.
Definition. Number sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. In this case the numbercalled the progression difference.
For an arithmetic progression, the following formulas are valid:
, (1)
Where . Formula (1) is called the formula of the general term of an arithmetic progression, and formula (2) represents the main property of an arithmetic progression: each term of the progression coincides with the arithmetic mean of its neighboring terms and .
Note that it is precisely because of this property that the progression under consideration is called “arithmetic”.
The above formulas (1) and (2) are generalized as follows:
(3)
To calculate the amount first terms of an arithmetic progressionthe formula is usually used
(5) where and .
If we take into account the formula (1), then from formula (5) it follows
If we denote , then
Where . Since , formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , from formula (5) it follows, What
Little known to most students is the property of arithmetic progression, formulated through the following theorem.
Theorem. If , then
Proof. If , then
The theorem has been proven.
For example , using the theorem, it can be shown that
Let's move on to consider typical examples of solving problems on the topic “Arithmetic progression”.
Example 1. Let it be. Find .
Solution. Applying formula (6), we obtain . Since and , then or .
Example 2. Let it be three times greater, and when divided by the quotient, the result is 2 and the remainder is 8. Determine and .
Solution. From the conditions of the example, the system of equations follows
Since , , and , then from the system of equations (10) we obtain
The solution to this system of equations is and .
Example 3. Find if and .
Solution. According to formula (5) we have or . However, using property (9), we obtain .
Since and , then from the equality the equation follows or .
Example 4. Find if .
Solution.According to formula (5) we have
However, using the theorem, we can write
From here and from formula (11) we obtain .
Example 5. Given: . Find .
Solution. Since, then. However, therefore.
Example 6. Let , and . Find .
Solution. Using formula (9), we obtain . Therefore, if , then or .
Since and then here we have a system of equations
Solving which, we get and .
Natural root of the equation is .
Example 7. Find if and .
Solution. Since according to formula (3) we have that , then the system of equations follows from the problem conditions
If we substitute the expressioninto the second equation of the system, then we get or .
Roots quadratic equation are And .
Let's consider two cases.
1. Let , then . Since and , then .
In this case, according to formula (6), we have
2. If , then , and
Answer: and.
Example 8. It is known that and. Find .
Solution. Taking into account formula (5) and the condition of the example, we write and .
This implies the system of equations
If we multiply the first equation of the system by 2 and then add it to the second equation, we get
According to formula (9) we have. In this regard, it follows from (12) or .
Since and , then .
Answer: .
Example 9. Find if and .
Solution. Since , and by condition , then or .
From formula (5) it is known, What . Since, then.
Hence , here we have a system of linear equations
From here we get and . Taking into account formula (8), we write .
Example 10. Solve the equation.
Solution. From given equation follows that . Let us assume that , , and . In this case .
According to formula (1), we can write or .
Since , then equation (13) has the only suitable root .
Example 11. Find the maximum value provided that and .
Solution. Since , then the arithmetic progression under consideration is decreasing. In this regard, the expression takes on its maximum value when it is the number of the minimum positive term of the progression.
Let us use formula (1) and the fact, that and . Then we get that or .
Since , then or . However, in this inequalitylargest natural number, That's why .
If the values of , and are substituted into formula (6), we get .
Answer: .
Example 12. Determine the sum of all two-digit natural numbers that, when divided by the number 6, leave a remainder of 5.
Solution. Let us denote by the set of all two-digit natural numbers, i.e. . Next, we will construct a subset consisting of those elements (numbers) of the set that, when divided by the number 6, give a remainder of 5.
Easy to install, What . Obviously , that the elements of the setform an arithmetic progression, in which and .
To establish the cardinality (number of elements) of the set, we assume that . Since and , it follows from formula (1) or . Taking into account formula (5), we obtain .
The above examples of problem solving can by no means claim to be exhaustive. This article is written based on the analysis modern methods solutions typical tasks on a given topic. For a more in-depth study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections school curriculum. – M.: Lenand / URSS, 2014. – 216 p.
3. Medynsky M.M. Full course elementary mathematics in problems and exercises. Book 2: Number sequences and progression. – M.: Editus, 2015. – 208 p.
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